Interesting simple property of concave function: the directional derivative must be positive only on some convex sets?

Question

Hi let $f$ be a concave function on $\mathbb R^n$. $x_1,x_2,y$ are points in $\mathbb R^n$.

Given that for all $i\in\{1,2\}$:

$(y-x_i)\cdot\nabla f(x_i)\geq 0$ . That is the directional derivative is given to be positive. The direction vector is $\vec{x_iy}$.

Let $z=\frac{x_1+x_2}{2}$. $z$ is the mid point of $x_1,x_2$.

Is it true that $(y-z)\cdot\nabla f(z)\geq 0$?

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I thought it is false but I cannot find a counterexample.

Answer 1

Hint

If $\ f(s,t)=-s^2-t^2\ $, then $\ f\ $ is a concave function. If \begin{align} x_1&=(1,1)\\ x_2&=(-1,1)\\ y&=\left(0,\frac{3}{2}\right) \end{align} what are \begin{align} (y-x_1)&\cdot\nabla f(x_1)\\ (y-x_2)&\cdot\nabla f(x_2)\ \ \text{ and}\\ (y-z)&\cdot\nabla f(z)\ ? \end{align}